How do you find critical points for G(x)= ^3sqrt(x²-x)?

1 Answer
Duncan D.
Apr 4, 2015

Hey there :)

The critical points of G(x) = (x^2-x)^(1/3) occur at x=1/2, x=0 and x=1 .

In order to find the critical points, you must take the derivative of G(x) . Using the chain rule, you should obtain G'(x)=(2x-1)/(3(x^2-x)^(2/3)) ( Try it! )

Critical points of a function are any points in the domain of the function where the value of the derivative of the function is 0 or undefined. (See http://en.wikipedia.org/wiki/Critical_point_%28mathematics%29 for more info!)

So for your function, we are looking for the x values such that G'(x) = 0 . This is (2x-1)/(3(x^2-x)^(2/3)) = 0 . Now the only way this is possible is if 2x-1 = 0 . This gives x = 1/2 .

But we must also see where G'(x) is undefined!
Now we can't have a zero denominator, so G'(x) is undefined if x^2-x = 0. Factoring, we get x(x-1) = 0 . This gives x = 0 and x=1 .

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